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THE BRAUER-MANIN OBSTRUCTION AND THE FIBRATION METHOD - LECTURE BY JEAN-LOUIS COLLIOT-THEL´ ENE`

These are informal notes on the lecture I gave at IU Bremen on July 14th, 2005. Steve Donnelly prepared a preliminary set of notes, which I completed.

1. The Brauer-Manin obstruction 2 For a scheme X, the Br X is H´et(X, Gm) ([9]). When X is a regular, noetherian, separated scheme, this coincides with the Azumaya Brauer group. If F ia a field, then Br(Spec F ) = Br F = H2(Gal(F /F ),F ×). Let k be a field and X a k-variety. For any field F containing k, each element A ∈ Br X gives rise to an “evaluation map” evA : X(F ) → Br F . For a number field k, class field theory gives a fundamental exact sequence L invv . 0 −−−→ Br k −−−→ all v Br kv −−−→ Q/Z −−−→ 0 That the composite map from Br k to Q/Z is zero is a generalization of the quadratic reciprocity law. Now suppose X is a projective variety defined over a number field k, and A ∈ Br X. One has the commutative diagram Q X(k) −−−→ v X(kv)   evA evA y y

Br k −−−→ ⊕v Br kv −−−→ Q/Z The second vertical map makes sense because, if X is projective, then for each fixed A, evA : X(kv) → Br kv is the zero map for all but finitely many v. Q Let ΘA : v X(kv) → Q/Z denote the composed map. Thus X(k) ⊆ ker ΘA for all A. Given a subgroup B ⊆ Br X, we write B \ X(Ak) := ker ΘA A∈B Q (where, for X projective, X(Ak) = v X(kv)). With this notation one has the inclusions Br X B X(k) ⊆ X(Ak) ⊆ X(Ak) ⊆ X(Ak) . In this way, each A ∈ Br X potentially obstructs the existence of k-rational points on X. Example: Let V/Q be the affine surface 0 6= x2 + y2 = (3 − t2)(t2 − 2) and let X ⊃ V be a smooth compactification. We take the element A ∈ Br X which is given on V by the 2 quaternion algebra (−1, 3 − t ). One finds that for all places v 6= 2, evA : X(kv) → Br kv is 1 identically zero; however at v = 2, evA is identically 2 . Therefore X(Q) must be empty. The surface V is a special case of a surface defined by an affine equation x2 − ay2 = P (t), with a ∈ k× and P (t) ∈ k[t] a polynomial of degree 4. A smooth projective model of such a surface is called a Chˆatelet surface. In 1984, Sansuc, Swinnerton-Dyer and I proved that Br X if X is a Chˆatelet surface, then the condition X(Ak) 6= ∅ is a necessary and sufficient 1 2 J-L. COLLIOT-THEL´ ENE`

Br X condition for the existence of a k-point. We actually showed that X(k) is dense in X(Ak) (under the diagonal embedding). One may wonder to which extent this result holds for other classes of varieties. That it could not extend to all varieties was foreseeable but it is only in 1999 that the first uncondi- Br X tional example of a smooth, projective, geometrically connected variety with X(Ak) 6= ∅ but X(k) = ∅ was produced (Skorobogatov, further work since then by Harari and Sko- robogatov, see [17]). For X a curve of genus one, if the Tate-Shafarevich group of the Jacobian of X is finite, Br X then X(Ak) 6= ∅ implies X(k) 6= ∅. This observation is due to Manin (1970). For curves of genus bigger than 1, quite surprisingly, it does not seem absurd to ask whether the same statement holds (see [18]).

2. Calculating the Brauer group Br X One would like to be able to compute X(Ak) . For this, a prerequisite is to compute the Brauer group Br X, or a least a system of representants of Br X/ Br k. Suppose char k is zero, and X/k is smooth, projective and geometrically connected. We write X := X ×k k, where k is an algebraic closure of k. 2.1. The “geometric” Brauer group. For computing Br X, we have an exact sequence

b2−ρ 3 0 → (Q/Z) → Br X → H (X, Z)tors → 0.

Here b2 is the second Betti number, which one computes by using either l-adic cohomology 2 2 H´et(X, Ql) for an arbitrary prime l or by using classical cohomology H (X ×k C, Q) if an embedding k ⊂ C is given. The integer ρ = rk NS X is the rank of the geometric N´eron-Severi group. The vanishing of b2 − ρ is equivalent to the vanishing of the coherent cohomology 2 3 group H (X,OX ). The group H (X, Z)tors is a finite group, which one computes either as 3 the direct sum over all primes l of the torsion in integral l-adic cohomology H´et(X, Zl) or as 3 the torsion in classical cohomology H (X ×k C, Z) if an embedding k ⊂ C is given. If X is a curve, or if X is birational to a projective space, then Br X = 0. Remarks 1. It is in general quite difficult to exhibit the Azumaya algebras on X corresponding to the divisible subgroup (Q/Z)b2−ρ. 2. When k is a number field, it is an open question whether the group of fixed points (Br X)Gal(k/k) is finite.

2.2. The “algebraic” Brauer group. Define Br1(X) := ker[Br X → Br X]. For comput- ing this group, we have the exact sequence

Gal(k/k) ∗ 1 ∗ 3 × 0 → Pic X → (Pic X) → Br k → Br1 X → H (k, Pic X) → H (k, k ) × where the maps marked with a ∗ are zero when X(k) is nonempty. The group H3(k, k ) is trivial when k is a number field (this is a nontrivial result from class field theory). There are cases where it is easy to explicitly compute the group H1(k, Pic X) but where it is difficult to lift a given element of that group to an explicit element of Br1(X) : Even if one knows a 3-cocycle is a 3-coboundary, it is not easy to write it down as an explicit Br X 3-coboundary. This may create difficulties for deciding whether a given X(Ak) is empty or not. Such a delicate situation arises in the study of diagonal cubic surfaces ([6], [13]). J-L. COLLIOT-THEL´ ENE` 3

To get a hold on H1(k, Pic X) one uses the exact sequence of Galois-modules 0 0 → PicX/k(k) → Pic X → NS X → 0 . 1 Here NS X is of finite type. If NS Xtors = 0, then H (k, NS X) is a finite group.

2.3. Curves. If X = C is a curve, then Br1 C = Br C (as noted above), and the above exact sequence reads 0 → Jac C(k) → Pic C → Z → 0. 1 Since H (k, Z) = 0, we thus have the exact sequence Gal(k/k) 1 1 (Pic C) → Z → H (k, Jac C(k)) → H (k, Pic C) → 0 which one may combine with the above long exact sequence. The group H1(k, Jac C(k)) 0 1 classifies principal homogeneous spaces under Jac C = PicC/k. The map Z → H (k, Jac C(k)) 1 sends 1 to the class of the principal homogeneous space PicC/k. If k is a number field, we thus have a surjective map from Br C to a quotient of H1(k, Jac C(k)). In practice, it is quite hard to lift an element of this quotient to an explicit element of Br C. Examples 1 1 1. If C = Pk, then the natural map Br k → Br Pk is an isomorphism. 2. If C is a smooth projective conic with no , we have an exact sequence 0 → Z/2 → Br k → Br C → 0

1 7→ [AC ] where [AC ] ∈ 2 Br k is the class corresponding to C. 2.4. Residues. Let A be a discrete valuation ring with field of fractions F and with residue 1 field κ of characteristic zero. There is a natural “residue map” Br F → H (κ, Q/Z) and an exact sequence 1 0 → Br A → Br F → H (κ, Q/Z) . Let k be a field of characteristic zero. Let X be a smooth, integral, k-variety with function field k(X). Given a closed integral subvariety Y ⊂ X of codimension 1, with function 1 field k(Y ), we may consider the residue map Br k(X) → H (k(Y ), Q/Z). One then has (Grothendieck) the exact sequence

M 1 0 → Br X → Br k(X) → H (k(Y ), Q/Z), Y where Y runs through all codimension 1 subvarieties of X as above. From the exactness of this sequence one deduces that Br X is a birational invariant for smooth, projective, integral k-varieties.

1 2.5. The projective line. Let us consider the special case X = Pk. As noted above, 1 Br Pk = Br k. The short exact sequence above thus reads 1 M 1 0 → Br k → Br k(P ) → H (kP , Q/Z), 1 P ∈Pk 1 where P runs through the closed points of Pk and kP is the residue field at such a point P . 4 J-L. COLLIOT-THEL´ ENE`

One may compute the cokernel of the last map : there is an exact sequence P N 1 L 1 P kP /k 1 0 −−−→ Br k −−−→ Br k( ) −−−→ 1 H (kP , / ) −−−−−−→ H (k, / ) −−−→ 0, P P ∈Pk Q Z Q Z

where NkP/k is the corestriction map. 2.6. Conic bundles over the projective line. Let X/k be a smooth, projective, geomet- 1 rically connected surface equipped with a morphism X → Pk whose generic fibre Xη is a 1 smooth conic over K = k(Pk) = k(t). After performing k-birational transformations one 1 may assume that for each closed point P ∈ Pk, the fibre XP is a conic over the residue field 1 1 kP , and that X → Pk is relatively minimal. There are finitely many points P ∈ Pk for which X is not smooth. At such a point P , there is a quadratic extension F /k over which X P √ P P P splits into a pair of transversal lines. Write FP = kP ( aP ). Let A ∈ Br K be the class of a quaternion algebra over K associated to the conic Xη/K, as in example 2 of section 2.3. We shall assume that A does not come from Br k. In the long exact sequence associated 1 1 1 to P in section 2.5, for each closed point P ∈ Pk, the residue δP (A) ∈ H (kP , Q/Z) lies in 1 H (FP /kP , Z/2) = Z/2. Using 2.3, 2.4 and 2.5, one shows that there is an exact sequence × ×2 0 → Br k → Br X → (⊕P (Z/2)P )/({δP (A)}) → k /k . 1 1 The last map sends the class of the element 1 ∈ (Z/2)P = H (FP /kP , Z/2) ⊂ H (kP , Q/Z) × ×2 to the class of NkP /k(aP ) ∈ k /k . In this situation one may give explicit generators for Br X/ Br k. They are given as the images under Br k(t) → Br k(X) of suitable linear combinations of elements of the shape × CoreskP /k(t−αP , βP ) ∈ Br k(t), where kP = k(αP ), βP ∈ kP , and (t−αP , βP ) is a quaternion algebra over the field kP (t). 2.7. Computing when no smooth projective model is available. One is often con- fronted with the following problem : given a smooth, affine, geometrically connected variety U over a field k, compute the Brauer group of a smooth compactification X of U without knowing a single such smooth compactification. The point as far as local to global problems are concerned is that it is only the Brauer group of smooth compactifications which natu- rally produces obstructions to the existence of rational points. A preliminary question is to compute H1(k, Pic X) (also a birational invariant of smooth, projective varieties). Assume U = T is a k-torus, i.e. an algebraic group which over k becomes isomorphic to a product of multiplicative groups. To such a k-torus there is associated its character group Tˆ (over k). This is a g-lattice (g being the Galois group of k over k). For X any smooth k-compactification of T , one has Y H1(k, Pic X) = ker[H2(g, Tˆ) → H2(h, Tˆ)], h where h ⊂ g runs through all closed pro-cyclic subgroups of g – as a matter of fact the computation of this kernel may be done after going over only to a suitable finite Galois extension of k. It seems hard to lift the elements of H1(k, Pic X) to explicit elements in Br X. The situation gets worse if X is a smooth compactification of a principal homogeneous space U under T . We have the same formula for H1(k, Pic X) as above, but in this case for k arbitrary J-L. COLLIOT-THEL´ ENE` 5

1 there is no reason why the map Br1 X → H (k, Pic X) should be surjective. If k is a number field, the map is surjective but lifting seems nevertheless very hard. Hence it seems difficult Br Xc to test the condition X(Ak) 6= ∅. Probably the simplest nontrivial example is the norm 1 torus T = R1 associated √ K/kGm √ 1 to a biquadratic extension K = k( a, b)/k. In this case one finds H (k, Pic X) = Z/2. The same result holds if X = Xc is a smooth compactification of a a principal homogeneous 1 space of RK/kGm, that is a variety U = Uc given by an equation NormK/k(z) = c for some × c ∈ k . If k is a number field, Uc has points in all completions of k, and Xc is a smooth Br Xc A compactification of U, then there exists some A ∈ Br Xc such that X(Ak) = X(Ak) . But how to compute such an A in a systematic fashion ? Br Xc The question is important, since in this case it is known that Xc(Ak) 6= ∅ implies Xc(k) 6= ∅. The latter statement is a general fact for principal homogeneous spaces of connected linear algebraic groups (Sansuc [15]), and it holds more generally for smooth compactifications of homogeneous spaces under connected linear algebraic groups, at least when the geometric stabilizer group is connected (Borovoi). √ √ Coming back to the case of equations NormK/k(z) = c, in the case K = k( a, b), Br Xc Sansuc [16] gives an algorithm to decide whether Xc(Ak) 6= ∅. It would be interesting to understand this algorithm better.

It is natural to study varieties which are given as the total space of a one-parameter family of principal homogeneous spaces. A special but already difficult case is that of varieties given by an affine equation

NormK/k(z) = P (t) where K/k is a finite field extension and P (t) a polynomial in one variable. In [5] the group H1(k, Pic X) for smooth projective models X of varieties defined by such an equation was computed for many cases, but it could not computed in all cases. See the questions raised at the end of section 2 of [5].

3. The fibration method Let k be a number field and X/k a smooth, projective, integral variety equipped with a 1 1 dominant k-morphism X → Pk whose generic fibre Xη is absolutely irreducible. For P ∈ P let XP denote the fibre above P (so XP is defined over kP ). 1 Question : Assume that the smooth fibres over k-points of Pk satisfy the , or at least that the Brauer-Manin obstruction to the Hasse principle for these fibres is the only one. Does it follow that the Brauer-Manin obstruction to the Hasse principle is the only obstruction for X ? It is not reasonable to expect a positive answer in the general case, for instance for pencils of curves of genus one with multiple fibres. Here is a context in which one may conjecture a positive answer : The generic fibre Xη is a smooth compactification of a (connected) homgeneous space of a connected algebraic group G/k(P1), and (i) the geometric stabilizer for this action is a connected group; 1 (ii) all fibres of X → Pk have at least one component of multiplicity one. (Condition (ii) is automatic when G is a connected linear algebraic group.) 6 J-L. COLLIOT-THEL´ ENE`

1 We call a closed point P ∈ P a good point if XP contains a multiplicity one component which is absolutely irreducible over kP . Set X δ := [kP : k] . bad P Theorem 3.1. Suppose that δ = 0, or that δ = 1 and f has a section over k. If the smooth fibres of f satisfy the Hasse principle, then X satisfies the Hasse principle. We have following theorem of D. Harari.

Theorem 3.2. [10] [11] Suppose δ ≤ 1 and f has a section over k. Also suppose Pic Xη is 1 torsion free, and Br Xη is finite. Assume that for all P ∈ P (k) for which XP is smooth,

BrXP XP (Ak) 6= ∅ =⇒ XP (k) 6= ∅ . Br X Then X(Ak) 6= ∅ implies X(k) 6= ∅ . The following corollary had been known for some time.

Corollary 3.3. Assume that whenever X is an intersection of two quadrics in P4, one has BrXP XP (Ak) 6= ∅ =⇒ XP (k) 6= ∅ . Then the Hasse principle holds for every smooth intersection of two quadrics in a projective space Pn with n ≥ 5. Recall Schinzel’s hypothesis, which asserts the following: given any finite set of irreducible polynomials {p1(x), . . . , pn(x)} ⊂ Z[x] satisfying the trivial necessary conditions, there are infinitely many positive integers m for which p1(m), . . . , pn(m) are all prime. The following result encompasses various earlier results (Sansuc and the speaker, Serre, Swinnerton-Dyer) : Theorem 3.4. [7] Assume Schinzel’s hypothesis. With notation and hypotheses as in the beginning of the section, suppose (1) the Hasse principle holds for smooth fibres of f : X → P1, and 1 (2) for all P ∈ Pk, there is a component ZP ⊆ XP of multiplicity one such that the algebraic closure of kP in k(ZP ) is abelian over kP . Br X Then X(Ak) 6= ∅ =⇒ X(k) 6= ∅ .

One wonders whether the same theorem holds when the Hasse principle hypothesis on the smooth fibres is replaced by the hypothesis that the Brauer-Manin obstruction to the Hasse principle for these is the only one. This question is open already when the generic fibre Xη satisfies the conditions of Theorem 3.2, for instance when the generic fibre is birational to a principal homogeneous space of a connected linear algebraic group. The abelianity condition in hypothesis (2) has so far prevented us from establishing such a result.

Here is a special case of this question. Let K/k be a finite field extension, P (t) ∈ k[t] a nonconstant polynomial of degree at least 2. Let X be a smooth projective model of the affine variety defined by the equation

NormK/k(z) = P (t). Br X Does X(Ak) 6= ∅ imply X(k) 6= ∅ ? J-L. COLLIOT-THEL´ ENE` 7

When K/k is cyclic, and the Schinzel hypothesis is granted, Theorem 3.4 yields a positive answer. When P (t) is separable and split of degree 2, and k = Q, the answer is in the affirmative. The proof ([12] [5]) combines descent theory and the circle method. In each of the following special cases, even at the expense of assuming the Schinzel hy- pothesis, the answer is not known : 1. The extension K/k is cubic but not Galois, and P is of degree 4 (or more). √ √ 2. K = k( a, b)/k is a biquadratic extension, and P is of degree 3 (or more). √ √ 3. K = k( a, b)/k is a biquadratic extension, and P (t) = c(t2 − a) for some c ∈ k×.

Over the last seven years, work has been done on varieties with a pencil of homogeneous spaces of curves of genus one ([8], [19], [20], [14], [21]). In all these works, one assumes that Tate-Shafarevich groups of elliptic curves are finite. Indeed, one takes the case of curves for granted. In many of these papers, one also assumes the Schinzel hypothesis. In the two papers [20] [14] however, one does without the Schinzel hypothesis; more precisely, in this case the Schinzel hypothesis is used in the only case when it has been established, that of one polynomial of degree one : in this case, this is Dirichlet’s theorem on primes in an arithmetic progression.

The reader will find more detailed introductions to the fibration method and its applica- tions in the surveys [1] [2] [3] and in the notes [4]. These reports do not cover the recent developments [14] [21].

References

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Jean-Louis Colliot-Th´el`ene CNRS, UMR 8628, Math´ematiques Universit´ede Paris-Sud Bˆatiment 425 F-91405 Orsay France

[email protected]